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Q.
If the distance between successive compressions and rarefactions is $1 \,m$ and velocity of sound is $360\,m \,s ^{- i }$, then the frequency is
Waves
Solution:
Distance between successive compressions and rarefactions is $\lambda / 2$.
$\therefore \,\,\,\,\frac{\lambda}{2}=1$ or $\lambda=2\, m$
As $v=v \lambda$
$ \therefore \,\,\,v=\frac{v}{\lambda}=\frac{360\, m\, s ^{-1}}{2 m }=180\, Hz$